When do you put double semicolons in F#? - f#

This is a stupid question. I've been reading a couple books on F# and can't find anything that explains when you put ;; after a statement, nor can I find a pattern in the reading. When do you end a statement with double semi-colons?

In the non-interactive F# code that's not supposed to be compatible with OCaml, you shouldn't need to ever need double semicolon. In the OCaml compatible mode, you would use it at the end of a top-level function declaration (In the recent versions, you can switch to this mode by using files with .ml extension or by adding #light "off" to the top).
If you're using the command-line fsi.exe tool or F# Interactive in Visual Studio then you'd use ;; to end the current input for F#.
When I'm posting code samples here at StackOverflow (and in the code samples from my book), I use ;; in the listing when I also want to show the result of evaluating the expression in F# interactive:
Listing from F# interactive
> "Hello" + " world!";;
val it : string = "Hello world!"
> 1 + 2;;
val it : int = 3
Standard F# source code
let n = 1 + 2
printf "Hello world!"
Sometimes it is also useful to show the output as part of the listing, so I find this notation quite useful, but I never explained it anywhere, so it's great that you asked!

Are you talking about F# proper or about running F# functions in the F# Interactive? In F# Interactive ;; forces execution of the code just entered. other than that ;; does not have any special meaning that I know of

In F#, the only place ;; is required is to end expressions in the interactive mode.
;; is left over from the transition from OCaml, where in turn it is left over from Caml Light. Originally ;; was used to end top-level "phrases"--that is, let, type, etc. OCaml made ;; optional since the typical module consists of a series of let statements with maybe one statement at the end to call the main function. If you deviate from this pattern, you need to separate the statements with ;;. Unfortunately, in OCaml, when ;; is optional versus required is hard to learn.
However, F# introduces two relevant modifications to OCaml syntax: indentation and do. Top-level statements have to go inside a do block, and indentation is required for blocks, so F# always knows that each top-level statement begin with do and an indent and ends with an outdent. No more ;; required.
Overall, all you need to know is that [O']Caml's syntax sucks, and F# fixes a lot of its problems, but maintains a lot of confusing backward compatibility. (I believe that F# can still compile a lot of OCaml code.)
Note: This answer was based on my experience with OCaml and the link Adam Gent posted (which is unfortunately not very enlightening unless you know OCaml).

Symbol and Operator Reference (F#)
http://msdn.microsoft.com/en-us/library/dd233228(v=VS.100).aspx
Semi Colon:
•Separates expressions (used mostly in verbose syntax).
•Separates elements of a list.
•Separates fields of a record.
Double Semi Colon:
http://www.ffconsultancy.com/products/fsharp_journal/free/introduction.html
Articles in The F#.NET Journal quote F# code as it would appear in an interactive session. Specifically, the interactive session provides a > prompt, requires a double semicolon ;; identifier at the end of a code snippet to force evaluation, and returns the names (if any) and types of resulting definitions and values.

I suspect that you have seen F# code written when #light syntax wasn't enabled by default (#light syntax is on by default for the May 2009 CTP and later ones as well as for Visual Studio 2010) and then ;; means the end of a function declaration.
So what is #light syntax? It comes with the #light declaration:
The #light declaration makes
whitespace significant. Allowing the
developer to omit certain keywords
such as in, ;, ;;, begin, and end.
Here's a code written without #light syntax:
let halfWay a b =
let dif = b - a in
let mid = dif / 2 in
mid + a;;
and becomes with light syntax:
#light
let halfWay a b =
let dif = b - a
let mid = dif / 2
mid + a
As said you can omit the #light declaration now (which should be the case if you're on a recent CTP or Visual Studio 2010).
See also this thread if you want know more on the #light syntax: F# - Should I learn with or without #light?

The double semi-colon is used to mark the end of a block of code that is ready for evaluation in F# interactive when you are typing directly into the interactive session. For example, when using it as a calculator.
This is rarely seen in F# because you typically write code into a script file, highlight it and use ALT+ENTER to have it evaluated, with Visual Studio effectively injecting the ;; at the end for you.
OCaml is the same.
Literature often quotes code written as it would appear if it had been typed into an interactive session because this is a clear way to convey not only the code but also its inferred type. For example:
> [1; 2; 3];;
val it : int list = [1; 2; 3]
This means that you type the expression [1; 2; 3] into the interactive session followed by the ;; denoting the end of a block of code that is ready to be evaluated interactively and the compiler replies with val it : int list = [1; 2; 3] describing that the expression evaluated to a value of the type int list.

The double semicolon most likely comes from OCaml since that is what the language is based on.
See link text
Basically its for historical purposes and you need it for the evaluator (repl) if you use it.

There is no purpose for double semi-colons (outside of F# interactive). The semi-colon, according to MSDN:
Separates expressions (used mostly
in verbose syntax).
Separates
elements of a list.
Separates
fields of a record.
Therefore, in the first instance, ;; would be separating the expression before the first semi-colon from the empty expression after it but before the second semi-colon, and separating that empty expression from whatever came after the second semi-colon (just as in, say C# or C++).
In the instance of the list, I suspect you'd get an error for defining an empty list element.
With regards to the record, I suspect it would be similar to separating expressions, with the empty space between the semi-colons effectively being ignored.
F# interactive executes the entered F# on seeing a double semi-colon.
[Updated to cover F# interactive - courtesy of mfeingold)

The history of the double semicolon can be traced back to the beginnings of ML when semicolons were used as a separator in lists instead of commas. In this ICFP 2010 - Tribute to Robin Milner video around 50:15 Mike Gordon mentions:
There was a talk on F# where someone asked "Why is there double semicolon on the end of F# commands?" The reason is the separator in lists in the original ML is semicolons, so if you wanted a list 1;2;3; and put it on separate lines- if you ended a line with semicolon you were not ending the phrase, so using double semicolon meant the end of the expression. Then in Standard ML the separator for lists became comma, so that meant you could use single semicolons to end lists.

Related

What is perl experimental feature, postderef?

I see use experimental 'postderef' being used in Moxie here on line 8. I'm just confused at what it does. The man pages for experimental are pretty vague too,
allow the use of postfix dereferencing expressions, including in interpolating strings
Can anyone show what you would have to do without the pragma, and what the pragma makes easier or possible?
What is it
It's simple. It's syntactic sugar with ups and downs. The pragma is no longer needed as the feature is core in 5.24. But in order for the feature to be supported in between 5.20 and 5.24, it had to be enabled with: use experimental 'postderef'. In the provided example, in Moxie, it's used in one line which has $meta->mro->#*; without it you'd have to write #{$meta->mro}.
Synopsis
These are straight from D Foy's blog, along with Idiomatic Perl for comparison that I've written.
D Foy example Idiomatic Perl
$gimme_a_ref->()->#[0]->%* %{ $gimme_a_ref->()[0] }
$array_ref->#* #{ $array_ref }
get_hashref()->#{ qw(cat dog) } #{ get_hashref() }{ qw(cat dog) }
These examples totally provided by D Foy,
D Foy example Idiomatic Perl
$array_ref->[0][0]->#* #{ $array_ref->[0][0] }
$sub->&* &some_sub
Arguments-for
postderef allows chaining.
postderef_qq makes complex interpolation into scalar strings easier.
Arguments-against
not at all provided by D Foy
Loses sigil significance. Whereas before you knew what the "type" was by looking at the sigil on the left-most side. Now, you don't know until you read the whole chain. This seems to undermine any argument for the sigil, by forcing you to read the whole chain before you know what is expected. Perhaps the days of arguing that sigils are a good design decision are over? But, then again, perl6 is still all about them. Lack of consistency here.
Overloads -> to mean, as type. So now you have $type->[0][1]->#* to mean dereference as $type, and also coerce to type.
Slices do not have an similar syntax on primitives.
my #foo = qw/foo bar baz quz quuz quuuz/;
my $bar = \#foo;
# Idiomatic perl array-slices with inclusive-range slicing
say #$bar[2..4]; # From reference; returns bazquzquuz
say #foo[2..4]; # From primitive; returns bazquzquuz
# Whizbang thing which has exclusive-range slicing
say $bar->#[2,4]; # From reference; returns bazquz
# Nothing.
Sources
Brian D Foy in 2014..
Brian D Foy in 2016..

Why is "do" allowed inside a function?

I noticed that the following code compiles and works in VS 2013:
let f() =
do Console.WriteLine(41)
42
But when looking at the F# 3.0 specification I can't find any mention of do being used this way. As far as I can tell, do can have the following uses:
As a part of loop (e.g. while expr do expr done), that's not the case here.
Inside computation expressions, e.g.:
seq {
for i in 1..2 do
do Console.WriteLine(i)
yield i * 2
}
That's not the case here either, f doesn't contain any computation expressions.
Though what confuses me here is that according to the specification, do should be followed by in. That in should be optional due to lightweight syntax, but adding it here causes a compile error (“Unexpected token 'in' or incomplete expression”).
Statement inside a module or class. This is also not the case here, the do is inside a function, not inside a module or a class.
I also noticed that with #light "off", the code doesn't compile (“Unexpected keyword 'do' in binding”), but I didn't find anything that would explain this in the section on lightweight syntax either.
Based on all this, I would assume that using do inside a function this way should not compile, but it does. Did I miss something in the specification? Or is this actually a bug in the compiler or in the specification?
From the documentation on MSDN:
A do binding is used to execute code without defining a function or value.
Even though the spec doesn't contain a comprehensive list of the places it is allowed, it is merely an expression asserted to be of type unit. Some examples:
if ((do ()); true) then ()
let x: unit = do ()
It is generally omitted. Each of the preceding examples are valid without do. Therefore, do serves only to assert that an expression is of type unit.
Going through the F# 3.0 specification expression syntax has do expr as a choice of class-function-or-value-defn (types) [Ch 8, A.2.5] and module-function-or-value-defn (modules) [Ch 10, A.2.1.1].
I don't actually see in the spec where function-defn can have more than one expression, as long all but the last one evaluate to unit -- or that all but the last expression is ignored in determining the functions return value.
So, it seems this is an oversight in the documentation.

F# Loading quotation data from an assembly - the explicitlyRegisterTopDefs function

I would like to understand how to retrieve the quotation from a top level function marked with [<ReflectedDefinition>] from an assembly.
It looks like this was done here: Tomas Petricek's blog: Quotation Visualiser Reloaded, but the code (at the very end of the article) makes a simple call to explicitlyRegisterTopDefs to retrieve the top level quoted definition.
I cannot seem to find this function in the latest version of the PowerPack or the F# compiler (I am working with .Net 4.0).
Lots of things happened to have changed since 2006 when the article was written, for example, the Microsoft.FSharp.Quotations.Raw was refactored, as you can see here.
Does anyone know how to capture these top level quotations with the latest versions of the PowerPack / compiler?
Thanks.
We did a lot of stuff like this WebSharper. Basically you do (no powerpack needed):
module QP = Quotations.Patterns
module QDP = Quotations.DerivedPatterns
[<ReflectedDefinition>]
let myFunc x = x + 1
match <# myFunc 1 #> with
| QP.Call(_, QDP.MethodWithReflectedDefinition d, _) ->
printfn "%A" d
| _ ->
printfn "ERROR"
I hope this helps with your scenario.
Note however that it has a ton of problems. Most grievous is that these active patterns throw exceptions from time to time. In addition, they are based on System.Reflection which slows things down enormously. Also, you have to account for unexpected things, like quotation currying not being resolved for you, certain constructor quotations failing, and so on.
For the upcoming WebSharper 2.4 I ended up rewriting the quotation loading code from scratch, using F# compiler sources as the definition of the binary format and avoiding System.Reflection, with great improvements in speed and reliability.

REBOL path operator vs division ambiguity

I've started looking into REBOL, just for fun, and as a fan of programming languages, I really like seeing new ideas and even just alternative syntaxes. REBOL is definitely full of these. One thing I noticed is the use of '/' as the path operator which can be used similarly to the '.' operator in most object-oriented programming languages. I have not programmed in REBOL extensively, just looked at some examples and read some documentation, but it isn't clear to me why there's no ambiguity with the '/' operator.
x: 4
y: 2
result: x/y
In my example, this should be division, but it seems like it could just as easily be the path operator if x were an object or function refinement. How does REBOL handle the ambiguity? Is it just a matter of an overloaded operator and the type system so it doesn't know until runtime? Or is it something I'm missing in the grammar and there really is a difference?
UPDATE Found a good piece of example code:
sp: to-integer (100 * 2 * length? buf) / d/3 / 1024 / 1024
It appears that arithmetic division requires whitespace, while the path operator requires no whitespace. Is that it?
This question deserves an answer from the syntactic point of view. In Rebol, there is no "path operator", in fact. The x/y is a syntactic element called path. As opposed to that the standalone / (delimited by spaces) is not a path, it is a word (which is usually interpreted as the division operator). In Rebol you can examine syntactic elements like this:
length? code: [x/y x / y] ; == 4
type? first code ; == path!
type? second code
, etc.
The code guide says:
White-space is used in general for delimiting (for separating symbols).
This is especially important because words may contain characters such as + and -.
http://www.rebol.com/r3/docs/guide/code-syntax.html
One acquired skill of being a REBOler is to get the hang of inserting whitespace in expressions where other languages usually do not require it :)
Spaces are generally needed in Rebol, but there are exceptions here and there for "special" characters, such as those delimiting series. For instance:
[a b c] is the same as [ a b c ]
(a b c) is the same as ( a b c )
[a b c]def is the same as [a b c] def
Some fairly powerful tools for doing introspection of syntactic elements are type?, quote, and probe. The quote operator prevents the interpreter from giving behavior to things. So if you tried something like:
>> data: [x [y 10]]
>> type? data/x/y
>> probe data/x/y
The "live" nature of the code would dig through the path and give you an integer! of value 10. But if you use quote:
>> data: [x [y 10]]
>> type? quote data/x/y
>> probe quote data/x/y
Then you wind up with a path! whose value is simply data/x/y, it never gets evaluated.
In the internal representation, a PATH! is quite similar to a BLOCK! or a PAREN!. It just has this special distinctive lexical type, which allows it to be treated differently. Although you've noticed that it can behave like a "dot" by picking members out of an object or series, that is only how it is used by the DO dialect. You could invent your own ideas, let's say you make the "russell" command:
russell [
x: 10
y: 20
z: 30
x/y/z
(
print x
print y
print z
)
]
Imagine that in my fanciful example, this outputs 30, 10, 20...because what the russell function does is evaluate its block in such a way that a path is treated as an instruction to shift values. So x/y/z means x=>y, y=>z, and z=>x. Then any code in parentheses is run in the DO dialect. Assignments are treated normally.
When you want to make up a fun new riff on how to express yourself, Rebol takes care of a lot of the grunt work. So for example the parentheses are guaranteed to have matched up to get a paren!. You don't have to go looking for all that yourself, you just build your dialect up from the building blocks of all those different types...and hook into existing behaviors (such as the DO dialect for basics like math and general computation, and the mind-bending PARSE dialect for some rather amazing pattern matching muscle).
But speaking of "all those different types", there's yet another weirdo situation for slash that can create another type:
>> type? quote /foo
This is called a refinement!, and happens when you start a lexical element with a slash. You'll see it used in the DO dialect to call out optional parameter sets to a function. But once again, it's just another symbolic LEGO in the parts box. You can ascribe meaning to it in your own dialects that is completely different...
While I didn't find any written definitive clarification, I did also find that +,-,* and others are valid characters in a word, so clearly it requires a space.
x*y
Is a valid identifier
x * y
Performs multiplication. It looks like the path operator is just another case of this.

Evaluate Quoted Expression in F#

I hope I haven't missed something obvious, but I've been playing with F# expressions and I want to evaluate quoted expressions on the fly. For example, I want write something like this:
let x = <# 2 * 5 #>
let y = transform x // replaces op_Multiply with op_Addition, or <# 2 + 5 #>
let z = eval y // dynamically evaluates y, returns 7
Is there a built-in F# method which can evaluate quoted expressions, or do I have to write my own?
I've implemented a reflection-based Quotation evaluator as part of Unquote (this is a new feature as of version 2.0.0).
> #r #"..\packages\Unquote.2.2.2\lib\net40\Unquote.dll"
--> Referenced '..\packages\Unquote.2.2.2\lib\net40\Unquote.dll'
> Swensen.Unquote.Operators.eval <# sprintf "%A" (1,2) #>;;
val it : string = "(1, 2)"
I've measured it to be up to 50 times faster than PowerPack's evaluator. This will, of course, vary by scenario. But Unquote is generally magnitudes faster than PowerPack at interpreting expressions.
It also supports many more expressions than PowerPack's evaluator, including VarSet, PropertySet, FieldSet, WhileLoop, ForIntegerRangeLoop, and Quote. In fact, Unquote's evaluator supports all quotation expressions except NewDelegate, AddressSet, and AddressOf all of which I plan on eventually supporting.
No, there's no built-in way to compile F# quotations. With the PowerPack LINQ you can convert SOME quotations to .NET System.Linq.Expressions.Expression, and use that to compile them.
Quotations were made to allow other interpretations of code, such as targeting SQL or a GPU card.
However, in posts on hubfs, it's been hinted at that this is a common request and will be looked at.
You can evaluate an F# quotation using the Eval extension member provided by the FSharp.PowerPack.Linq DLL as follows:
#r "FSharp.PowerPack.Linq.dll"
open Linq.QuotationEvaluation
let f = <#2 + 3#>
f.Eval()
Note that you must open the Linq.QuotationEvaluation namespace to make this extension member available.
There is also a Compile extension member that returns a suspension but it does not appear to improve performance.
Updated in 2016
The Evaluate extension method can now be found in NuGet package FSharp.Quotations.Evaluator
#r "../packages/FSharp.Quotations.Evaluator.1.0.7/lib/net40/FSharp.Quotations.Evaluator.dll"
open FSharp.Quotations.Evaluator
let f = <# 2 + 3 #>
f.Evaluate()
I think the quotations has got a .eval()-method.

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